Q15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 11–18, find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{sect}\mathbf{-}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$

Step-by-Step Solution

Answer

The general solution is $y\left(t\right)=c_1\cos t+c_2\sin t+3\ln \left|\cos t\right|\cos t-t^2+3+3t\sin t$.

Step 1: Find a particular solution.

The homogenous equation is $r^2+1=0$.

 

Two independent solutions are $r=\pm i$.

 

Then $y_1=\cos t,v_2=\sin t$

$y_h\left(t\right)=c_1\cos t+c_2\sin t$

 

The particular solution is $y_p=v_1\left(t\right)cost+v_2\left(t\right)sint$.

Step 2: Evaluate v 1     and     v 2

Here $y_p=v_1\left(t\right)\cos t+v_2\left(t\right)\sin t$

Here $f=3\sec t-t^2+1$ and $a=1$

 

And referring to (9) $y\left(t\right)=c_1{e}^{\alpha t}\cos \beta t{c}_2{e}^{\alpha t}\sin \beta t$ and solve the system by derivative then;

$\begin{array}{rcl}{v}_1\text{'}\cos t+v_2\text{'}\sin t& =& 0\\ -v_1\text{'}\sin t+v_2\cos t& =& \frac{f}{a}\\ -v_1\text{'}\sin t+v_2\cos t& =& 3\sec t-t^2+1\end{array}$

Step 3: Find v 1 ' and v 1

$\begin{array}{rcl}{v}_1\text{'}& =& \frac{-f\left(t\right)v_2\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]}\\ & =& \frac{-\left(3\sec t-t^2+1\right).\sin t}{\left[{\cos }^{2}t+{\sin }^{2}t\right]}\\ & =& -\left(3\sec t-t^2+1\right).\sin t\end{array}$

Now integrating this;

$\begin{array}{rcl}{v}_1\left(t\right)& =& \int -\left(3\sec t-t^2+1\right).\sin tdt\\ & =& 2\ln \left|\cos t\right|-t^2\cos t++2t\sin t+3\cos t+C\end{array}$

Step 4: Determine v 2 ' and v 2

$\begin{array}{rcl}{v}_2\text{'}& =& \frac{f\left(t\right)y_1\left(t\right)}{a\left[y_1\left(t\right)y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right)y_2\left(t\right)\right]}\\ & =& \frac{-\left(3\sec t-t^2+1\right).\cos t}{\left[{\cos }^{2}t+{\sin }^{2}t\right]}\\ & =& \left(3\sec t-t^2+1\right).\cos t\end{array}$

Integrate this.

 $\begin{array}{rcl}{y}_2\left(t\right)& =& \int \left(3\sec t-t^2+1\right).\cos tdt\\ & =& 3t-t^2\sin t-2t\cos t+3\sin t+C\end{array}$

 

Thus, the particular solution is:

$\begin{array}{rcl}{y}_p& =& \left(2\ln \left|\cos t\right|-t^2\cos t++2t\sin t+3\cos t+C\right)\cos t+\left(3t-t^2\sin t-2t\cos t+3\sin t+C\right)\sin t\\ y_p& =& 3\ln \left|\cos t\right|\cos t-t^2+3+3t\sin t\end{array}$ 

Therefore, the general solution is:

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_{1}cost+c_{2}sint+3ln\left|cost\right|cost-t^2+3+3tsint \end{gathered}$$